Pair Annihilation into Gamma Rays

There are three essentially different kinds of pair annihilations. The first kind is the annihilation of a free positron with a free electron in relative motion towards each other; we shall call this the annihilation of free electron-positron pairs and shall discuss it here. Since the probability for the occurrence of this process increases with decreasing relative velocity, it will compete with the process of capture into the bound positronium state, which in turn can decay into photons. This positronium annihilation most likely occurs from one of the lowest energy levels (S-state) and will always be governed by selection rules. Finally, the third kind of annihilation takes place in the presence of an external field, e.g. the Coulomb field of a nucleus. This pair annihilation in an external field is of importance when positrons annihilate with tightly bound electrons whose binding to the nucleus cannot be neglected. This latter process is the only one in which energy-momentum conservation does not exclude one-photon annihilation, i.e. pair annihilation with the emission of only one photon.

Consider the process of annihilation of an electron-positron pair into two gamma rays, as shown in figure 7.9. This is the lowest order in $e^2$ in which this process can occur, since pair annihilation to a single photon cannot conserve energy and momentum: $(p_-+p_+)^2 = 2m^2 + 2E_-E_+(1-\beta_-\beta_+\cos\theta) > 0$ but $k^2 = 0$. The diagrams can be view as Compton scattering turn on its side.

Figure 7.9: Feynman diagrams for pair annihilation into gamma rays.

Using our previously determined Feynman rules, we can write down the relevant $S$-matrix element in momentum space from inspection of the diagram in figure 7.9:

$$S_{fi} = \frac{e^2}{V^2} \sqrt{\frac{m^2}{E_+E_-2k_12k_2}} (2\pi)^4 \delta^4(k_1+k_2-p_+-p_-) \overline{v}(p_+,s_+)$$

$$\left[ (-i\not{\;\!\!\!\varepsilon}_2) \frac{i}{\not{\;\!\!\!p}_-... ...}_- -\not{\;\!\!\!k}_2 -m} (-i\not{\;\!\!\!\varepsilon}_2) \right] u(p_-,s_-) .$$ (7.265)

We notice that the $S$-matrix element is symmetric under interchange of the two photons, which is required by Bose statistics. Both graphs must be included in order to ensure this required symmetry.

Compared to Compton scattering the substitutions

$$\textrm{Compton} \leftrightarrow \textrm{Pair Annihilation}$$

$$\epsilon, k \leftrightarrow \epsilon_1, -k_1$$ (7.266)

$$\epsilon^\prime, k^\prime \leftrightarrow \epsilon_2, +k_2$$ (7.267)

$$p_i, s_i \leftrightarrow p_-, s_-$$ (7.268)

$$p_f, s_f \leftrightarrow -p_+, s_+$$ (7.269)

transform the two amplitudes into each other. This is an example of a general substitution rule which is valid to arbitrary order and which relates processes of the type $A+B\rightarrow C+D$ to the process $A+\overline{C}\rightarrow\overline{B}+D$, where $\overline{B}$ denotes the antiparticle to $B$, etc.

By familiar steps, we proceed from the matrix element to a differential cross-section

$$d\sigma = \frac{\vert S_{fi}\vert^2}{VT} \frac{V}{\vert\vec{J}_{inc}\vert} \frac{Vd^3k_1}{(2\pi)^3} \frac{Vd^3k_2}{(2\pi)^3}$$

$$= \frac{e^4}{(2\pi)^2} \int \frac{m^2}{E_+E_-\vert\vec{v}_+-\vec{v}... ...{fi}\vert^2 \frac{d^3k_1}{2k_1} \frac{d^3k_2}{2k_2} \delta^4(k_1+k_2-p_+-p_-) .$$ (7.270)

For an electron at rest

$$d\sigma = \frac{e^4}{(2\pi)^2} \int \frac{m}{E_+\beta_+} \ve... ...d^3k_1}{2k_1} \frac{d^3k_2}{2k_2} \delta^4(k_1+k_2-p_+-p_-) ,$$ (7.271)

where $\beta_+=p_+/E_+$ is the incident positron velocity. The invariant amplitude is

$$\mathcal{M}_{fi} = \overline{v}(p_+,s_+) \left[ (-i\not{\;\!\!\!\varepsilon}_2) \fra... ...}_- - \not{\;\!\!\!k}_2 - m} (-i\not{\;\!\!\!\varepsilon}_2) \right] u(p_-,s_-)$$

$$= i\overline{v}(p_+,s_+) \left[ \not{\;\!\!\!\varepsilon}_2 \frac{\... ...\;\!\!\!k}_2 + m}{2p_-\cdot k_2} \not{\;\!\!\!\varepsilon}_2 \right] u(p_-,s_-)$$

$$= -i\overline{v}(p_+,s_+) \left[ \frac{\not{\;\!\!\!\varepsilon}_2\... ...ot{\;\!\!\!k}_2\not{\;\!\!\!\varepsilon}_2}{2p_-\cdot k_2} \right] u(p_-,s_-) ,$$ (7.272)

where we have used $(\not{\;\!\!\!p}_-+m)\not{\;\!\!\!\varepsilon}u(p_-,s_-) = \not{\;\!\!\!\varepsilon}(-\not{\;\!\!\!p}_-+m) u(p_-,s_-) = 0$. For an unpolarized positron incident on an unpolarized electron, we average over the initial spins $s_-$ and $s_+$. Representing the quantity in square brackets as $\Gamma$, we can write

$$\frac{1}{4}\sum_{s_-,s_+}\vert\mathcal{M}_{fi}\vert^2 = \frac{1}{4}\sum_{s_-,s_+}\overline{v}(p_+,s_+)_\alpha \Gamma_{\al... ...eta \overline{u}(p_-,s_-)_\delta \overline{\Gamma}_{\delta\rho} v(p_+,s_+)_\rho$$

$$= \frac{1}{4}\sum_{s_+}\overline{v}(p_+,s_+)_\alpha \Gamma_{\alpha\... ...}_-+m}{2m} \right)_{\beta\delta} \overline{\Gamma}_{\delta\rho} v(p_+,s_+)_\rho$$

$$= -\frac{1}{4}\left( \frac{m-\not{\;\!\!\!p}_+}{2m} \right)_{\rho\a... ...c{\not{\;\!\!\!p}_-+m}{2m} \right)_{\beta\delta} \overline{\Gamma}_{\delta\rho}$$

$$= -\frac{1}{4}\textrm{Tr} \left[ \frac{m-\not{\;\!\!\!p}_+}{... ...2\not{\;\!\!\!\varepsilon}_1}{2p_-\cdot k_2} \right) \right] .$$ (7.273)

The unpolarized cross-section becomes

$$d\overline{\sigma} = -\frac{e^4}{(2\pi)^2} \int \frac{m}{E_+\beta_+}$$

$$\cdot \frac{1}{4} \textrm{Tr} \left[ \frac{m-\not{\;\!\!\!p}_+}{2m} \le... ...}_2\not{\;\!\!\!k}_2\not{\;\!\!\!\varepsilon}_1}{2p_-\cdot k_2} \right) \right]$$

$$\cdot \frac{d^3k_1}{2k_1} \frac{d^3k_2}{2k_2} \delta^4(k_1+k_2-p_--p_+) .$$ (7.274)

Notice the minus sign which comes for our normalization of the positron spinors. The simplified form of the matrix element is due to the choice of transverse gauge $\epsilon_1\cdot p_- = \epsilon_2\cdot p_- = 0$ for the laboratory frame of reference and is the same gauge used in the Compton scattering calculation. We can thus obtain the trace directly from our previous result using the substitutions above:

$$d\overline{\sigma} = \frac{e^4}{(2\pi)^2} \int \frac{m}{E_+\beta_+} \frac{-1}{4} \frac... ...epsilon_1\cdot\epsilon_2)^2 - 2 \right] \frac{d^3k_1}{2k_1} \frac{d^3k_2}{2k_2}$$

$$\cdot \delta^4(k_1+k_2-p_--p_+)$$

$$= \frac{\alpha^2}{2m} \int \frac{1}{p_+} \left[ \frac{k_2}{k_1} + \... ...)^2 \right] \frac{d^3k_1}{2k_1} \frac{d^3k_2}{2k_2} \delta^4(k_1+k_2-p_--p_+) .$$ (7.275)

It remains to reduce the $\delta$-function for laboratory kinematics:

$$\int \frac{d^3k_1}{2k_1} \frac{d^3k_2}{2k_2} \delta^4(k_1+k_2-p_+-p_-)$$

$$= \int \frac{d^3k_1}{2k_1} \int_{-\infty}^\infty d^4k_2 \delta^4(k_1+k_2-p_+-p_-) \delta(k_2^2) \theta(k_2^0)$$

$$= \int \frac{d^3k_1}{2k_1} \delta^4[(p_++p_--k_1)^2] \theta(E_++E_--k_1)$$

$$= \int_0^\infty \frac{1}{2} k_1dk_1d\Omega_{k_1} \delta[(p_+ + p_-)^2 -2k_1\cdot(p_+ + p_-)] \theta(E_+ + E_- - k_1)$$ (7.276)

$$= \frac{d\Omega_{k_1}}{2} \int_0^{E_++m} k_1 dk_1 \delta[2m^2 + 2mE_+ -2k_1(m + E_+ -p_+\cos\theta)]$$ (7.277)

$$= \frac{1}{4} \frac{m(m+E_+)}{[m + E_+ - p_+\cos\theta]^2} d\Omega_{k_1} .$$ (7.278)

The $\delta$-functions allows us to write ( $z\equiv\cos\theta$, $\beta = p_+/E_+$, $\gamma = E_+/m$)

$$\frac{k_1}{m} = \frac{(m + E_+)}{m + E_+ - p_+\cos\theta} = \frac{1/\gamma+1}{1/\gamma+1-\beta z} = \frac{1+\gamma}{1+\gamma(1-\beta z)} ,$$ (7.279)

$$\frac{k_2}{m} = 1 + \frac{E_+}{m} -\frac{k_1}{m} = \frac{(1+\gamma)\gamma(1 - \beta z)}{1+\gamma(1-\beta z)},$$ (7.280)

$$\frac{k_2}{k_1} = \gamma(1-\beta z) .$$ (7.281)

Combining the above expressions gives

$$\frac{d\overline{\sigma}}{d\Omega_{k_1}} = \frac{\alpha^2}{8... ...+ \frac{k_2}{k_1} + 2 -4(\epsilon_1\cdot\epsilon_2)^2\right] .$$ (7.282)

The angle between the two photons, $\Theta$, can be found from

$$(k_1+k_2)^2 = (p_-+p_+)^2$$

$$2k_1\cdot k_2 = 2m^2 + 2mE$$

$$2k_1k_2(1-\cos\Theta) = 2m(m+E)$$

$$1-\cos\Theta = \frac{m(m+E)}{k_1k_2} = m\frac{k_1+k_2}{k_1k_2} = m\left( \frac{1}{k_2} + \frac{1}{k_2} \right) .$$ (7.283)

This leads to

$$\frac{d\overline{\sigma}}{d\Omega_{k_1}} = \frac{\alpha^2}{8... ...+ \frac{k_2}{k_1} + 2 -4(\epsilon_1\cdot\epsilon_2)^2\right] .$$ (7.284)

The sum over the final polarization states in the rest system gives (cf. equation 7.253)

$$\sum (\epsilon_1\cdot\epsilon_2)^2 = 1 + \cos^2\Theta$$ (7.285)

and a factor of 4 for those terms which are independent of $\epsilon_1$ and $\epsilon_2$, where $\Theta$ is the angle between the photons. We now have

$$\frac{d\overline{\sigma}}{d\Omega_{k_1}} = \frac{\alpha^2}{2m^2\beta\gamma} \left(\frac{k_1}{k_2}\right) \fr... ...-\cos\Theta} \left[ \frac{k_1}{k_2} + \frac{k_2}{k_1} + 1 -\cos^2\Theta \right]$$

$$= \frac{\alpha^2}{2m^2\beta\gamma} \left(\frac{k_1}{k_2}\right) \fr... ...(k_1+k_2) \left( \frac{1}{k_2} + \frac{1}{k_1}\right) - 1 -\cos^2\Theta \right]$$

$$= \frac{\alpha^2}{2m^2\beta\gamma} \left(\frac{k_1}{k_2}\right) \frac{1}{1-\cos\Theta} \left[ (1+\gamma) ( 1-\cos\Theta) - (1 +\cos^2\Theta) \right]$$

$$= \frac{\alpha^2}{2m^2\beta\gamma} \frac{k_1}{k_2} \left[ 1 + \gamma - \frac{1 +\cos^2\Theta}{1-\cos\Theta} \right] .$$ (7.286)

We now express the cross-section in terms of the energies of the two photons. Simple kinimatics gives

$$\cos\Theta = 1 -\frac{m}{k_1} -\frac{m}{k_2}$$

$$\cos^2\Theta = 1 + \left(\frac{m}{k_1}\right)^2 + \left(\frac{m}{k_2}\right)^2 -2\frac{m}{k_1} -2\frac{m}{k_2} + 2\frac{m}{k_1}\frac{m}{k_2}$$ (7.287)

$$\frac{1+\cos^2\Theta}{1-\cos\Theta} = \frac{m}{k_1+k_2} \left[ 2\frac{k_1}{m}\frac{k_2}{m} + \frac{k_2}{k_1} + \frac{k_1}{k_2} -2\frac{k_1}{m} -2\frac{k_2}{m} + 2 \right]$$

$$= \frac{1}{1+\gamma} \left[ \frac{2\gamma(1+\gamma)^2(1-\beta z)}{[... ...)]^2} + \gamma(1-\beta z)+ \frac{1}{\gamma(1-\beta z)} -2(1+\gamma) + 2 \right]$$

$$= \frac{1}{1+\gamma} \left[ \frac{2\gamma(1+\gamma)^2(1-\beta z)}{[... ...]^2} + \frac{[1+\gamma(1-\beta z)]^2}{\gamma(1-\beta z)} -2(1+\gamma) \right] .$$ (7.288)

The differential cross-section in terms of photons energies becomes

$$\frac{d\overline{\sigma}}{d\Omega_{k_1}} = \frac{\alpha^2}{2... ...gamma(1+\gamma)(1-\beta z)}{[1+\gamma(1-\beta z)]^2} \right] .$$ (7.289)

For a total cross-section, we sum $d\overline{\sigma}/d\Omega_{k_1}$ over final photon polarizations and integrate over the solid angle. Since the final state contains two identical particles, one of the photons emerges in $d\Omega_{k_1}$; because of their indistinguishability, this can be either of the two photons. If we were to integrate $d\sigma/d\Omega_{k_1}$ over the entire $4\pi$ solid angle, we would be counting each distinguishable state exactly twice. We therefore take one-half of this integral in forming a total cross-section

$$\overline{\sigma} = \frac{1}{2} \int \frac{d\overline{\sigma}}{d\Omega_{k_1}} d\Omega_{k_1} .$$ (7.290)

We now integrate the differential cross-section using:

$$\int_{-1}^1\frac{dz}{[1+\gamma(1-\beta z)]^2} = \frac{1}{1+\gamma} ,$$ (7.291)

$$\int_{-1}^1\frac{dz}{(1-\beta z)^2} = 2\gamma^2 ,$$ (7.292)

$$\int_{-1}^1\frac{dz}{1-\beta z} = -\frac{1}{\beta}\ln\frac{1-\beta}{1+\beta} .$$ (7.293)

Using these integrals we have

$$\sigma = \frac{\pi\alpha^2}{2m^2\beta\gamma^2} \left[ -\frac{\gamma+3}{\be... ...\gamma)} \ln \frac{1-\beta}{1+\beta} -\frac{2\gamma}{1+\gamma} -2\gamma \right]$$

$$= \frac{\pi\alpha^2}{2m^2\beta^2\gamma^2(\gamma+1)} \left[ (\gamma^2 + 4\gamma + 1) \ln\frac{1+\beta}{1-\beta} - 2\beta\gamma (\gamma + 3) \right]$$

$$= \frac{\pi\alpha^2}{m^2\beta^2\gamma(\gamma+1)} \left[ \left(\gamm... ...}{\gamma}\right) \ln\sqrt{\frac{1+\beta}{1-\beta}} - \beta (\gamma + 3) \right]$$

$$= \frac{\pi\alpha^2}{m^2\beta^2\gamma(\gamma+1)} \left[ \left( \gam... ...{\gamma} \right) \ln(\gamma + \sqrt{\gamma^2-1}) - \beta (\gamma + 3) \right] .$$ (7.294)

For incident positrons of low energy, $\gamma\rightarrow 1$ and $\beta\rightarrow 0$ giving

$$\overline{\sigma} \approx \frac{\pi\alpha^2}{2m^2\beta^2} \left[ (6-1/2\beta^4+\cdots) (\beta -1/4\beta^4 + \cdots) - 4\beta -1/2\beta^3 \cdots \right]$$

$$\approx \frac{\pi\alpha^2}{2m^2\beta^2} \left[ 2\beta -1/2\beta^3 + \cdots \right]$$

$$= \frac{\pi\alpha^2}{m^2\beta} \left[ 1 + \mathcal{O}(\beta) \right], \quad \textrm{for} \quad \beta\ll 1.$$ (7.295)

In the extreme relativistic limit, $\gamma\gg 1$ and $\beta\rightarrow 1$ giving

$$\overline{\sigma} = \frac{\pi\alpha^2}{m^2\gamma^2} \left[ \left( \gamma + 4 \cdots + \right) \ln(\gamma + \gamma(1 -\cdots)) -\gamma \right]$$

$$= \frac{\pi\alpha^2}{m^2\gamma} \left[ \ln 2\gamma - 1 + \frac{4}{\gamma} \ln(2\gamma) + \cdots \right]$$

$$= \frac{\pi\alpha^2}{mE_+} \left[ \ln \frac{2E_+}{m} + \mathcal{O} ... ...c{m}{E_+}\ln\frac{E_+}{m} \right) \right] \quad \textrm{for} \quad \gamma\gg 1.$$ (7.296)